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# If |2z| – 1 ≥ 2, which of the following graphs could be a nu

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Kudos [?]: 314 [0], given: 1279

If |2z| – 1 ≥ 2, which of the following graphs could be a nu [#permalink]  21 Jun 2017, 13:17
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If |2z| – 1 ≥ 2, which of the following graphs could be a number line representing all the possible values of z?

[Reveal] Spoiler: OA

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Senior Manager
Joined: 03 Sep 2017
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Kudos [?]: 124 [1] , given: 64

Re: If |2z| – 1 ≥ 2, which of the following graphs could be a nu [#permalink]  12 Oct 2017, 01:47
1
KUDOS
The equation is solved as

for x >= 0, 2z-1\geq 2 i.e. z\geq \frac{3}{2}

for x <= 0, -2z-1\geq 2 i.e. z\leq -\frac{3}{2}

Thus, the value of x are x\leq -\frac{3}{2} and x \geq \frac{3}{2}.

The line number is A because 3/2 is greater than 1 and -3/2 is less than -1
Intern
Joined: 12 Oct 2017
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Kudos [?]: 11 [1] , given: 3

Re: If |2z| – 1 ≥ 2, which of the following graphs could be a nu [#permalink]  12 Oct 2017, 09:19
1
KUDOS
Solutions :-

There Can be multiple ways to think over this problem but whenever we are finding answer from the given choices try to eliminate the wrong choice instead of finding the right one ( this is very useful in most of the cases )

In above problem, from the given choices we need to check the value for z=1 or -1 as for both |z| = 1

for z=1
equation is "1>=2"
which is not correct
so |z| should be greater than 1
or z should be greater than 1 or less than -1

if we check the choices, Choice "B,C,D,E" is eliminated as all are wrong so we are left with only Choice "A"

GRE Instructor
Joined: 10 Apr 2015
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Kudos [?]: 450 [1] , given: 3

Re: If |2z| – 1 ≥ 2, which of the following graphs could be a nu [#permalink]  12 Oct 2017, 15:17
1
KUDOS
Expert's post
Carcass wrote:

If |2z| – 1 ≥ 2, which of the following graphs could be a number line representing all the possible values of z?

Another approach is to TEST SOME values.

When I scan the answer choices, I see that some have 0 as part of the solution, and others don't.
So, let's see what happens when z = 0

Plug z = 0 into original equation to get: |2(0)| – 1 ≥ 2
Evaluate to get: |0| – 1 ≥ 2
We get: 0 – 1 ≥ 2
And then: -1 ≥ 2
NOT TRUE.
So, z = 0 is NOT a solution.
Therefore, we'll ELIMINATE D and E, since they include z=0 as a solution.

Among the three remaining answer choices, two have have 1 as part of the solution, and one doesn't.
So, let's see what happens when z = 1
Plug z = 1 into original equation to get: |2(1)| – 1 ≥ 2
Evaluate to get: |2| – 1 ≥ 2
We get: 2 – 1 ≥ 2
And then: 1 ≥ 2
NOT TRUE.
So, z = 1 is NOT a solution.
Therefore, we'll ELIMINATE B and C, since they include z=1 as a solution.

By the process of elimination, we're left with A, the correct answer.

Cheers,
Brent
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Brent Hanneson – Founder of greenlighttestprep.com

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Re: If |2z| – 1 ≥ 2, which of the following graphs could be a nu   [#permalink] 12 Oct 2017, 15:17
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